题目如图,BE=CFBE=CFBE=CF,AE⊥BCAE\bot BCAE⊥BC,DF⊥BCDF\bot BCDF⊥BC,要根据"HLHLHL"证明Rt△ABERt\triangle ABERt△ABE≌△DCF,\triangle DCF,△DCF,则还需添加一个条件是( )A.ABABAB∥CDCDCDB.AE=DFAE=DFAE=DFC.AB=CDAB=CDAB=CDD.∠B=∠D\angle B=\angle D∠B=∠D知识点:直角三角形全等的判定章节:图形的性质 / 三角形